The discrete random variable \(\mathrm{X}\) can take all possible integer values from 1 to \(\mathrm{k}\), each with a probability \(\frac{1}{\mathrm{k}}\), then its variance is

\(\begin{aligned} & \mathrm{E}(\mathrm{X})=\sum_{\mathrm{i}=1}^{\mathrm{k}} x_{\mathrm{i}} \mathrm{p}_{\mathrm{i}}=\frac{1}{\mathrm{k}}+\frac{2}{\mathrm{k}}+\ldots+\frac{\mathrm{k}}{\mathrm{k}}=\frac{\mathrm{k}+1}{2} \\ & \mathrm{E}\left(\mathrm{X}^2\right)=\sum_{\mathrm{i}=1}^{\mathrm{k}} x_{\mathrm{i}}^2 \mathrm{p}_{\mathrm{i}}=\frac{1^2+2^2+3^2+\ldots+\mathrm{k}^2}{\mathrm{k}} \\ & =\frac{(\mathrm{k}+1)(2 \mathrm{k}+1)}{6} \\ & \therefore \quad \text { Variance }=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\ & =\frac{2 \mathrm{k}^2+3 \mathrm{k}+1}{6}-\frac{\mathrm{k}^2+2 \mathrm{k}+1}{4} \\ & \end{aligned}\)
\(\begin{aligned} & =\frac{4 \mathrm{k}^2+6 \mathrm{k}+2-3 \mathrm{k}^2-6 \mathrm{k}-3}{12} \\ & =\frac{\mathrm{k}^2-1}{12}\end{aligned}\)